Hi i have to resolve AX=0 in matlab. The problem is that I don't have the A coefficients I need to extract them from q(x)f(x) - q(x) = 0, a complex expression. In this system the variables are a0 a1 a2 b0 b1 b2, since I going to evaluate the expression for a given set of points x, p(x) = a0 + a1*x + a2*x^2 and q(x) = b0 + b1*x + b2*x^2 and f(x) is some function. So i got a system of 6 variables and the number of equation is the quantity of points. My question is how i extract the coefficients of the A matrix including the 0 for any variable? I have been trying several ways but nothing. Do I have to manually copy the variables coefficients for all the given points (x, f(x))? Please i like the thing well done help me in case there is a solution even if it is big. just guide me.
So, I guess you want to solve q(x)f(x) - p(x) = 0 (where q,p is quadric and f arbitary function, respect to x), not AX=0. Note: AX=0 is linear equation while q(x)f(x) - p(x) = 0 not necessarily. If so, it may be impossible to "extract coefficients" and what you need is nonlinear equation solver, check out: fsolve function.
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Working in MATLAB. I have the following equation:
S = aW + bX + cY + dZ
where S,W,X,Y, and Z are all known n x 1 vectors. I am trying to fit the data of S with a linear combination of the basis vectors W,X,Y, and Z with the constraint of the constants (a,b,c,d) being greater than 0. I have managed to do this in Excel's solver, and have attempted to figure it out on MATLAB, being directed towards functions like fmincon, but I am not overly familiar with MATLAB and feel I am misunderstanding the use of fmincon.
I am looking for help with understanding fmincon's use towards my problem, or redirection towards a more efficient method for solving.
Currently I have:
initials = [0.2 0.2 0.2 0.2];
fun = #(x)x(1)*W + x(2)*X + x(3)*Y + x(4)*Z;
lb = [0,0,0,0];
soln = fmincon(fun,initials,data,b,Aeq,beq,lb,ub);
I am receiving an error stating "A must have 4 column(s)". Where A is referring to my variable data which corresponds to my S in the above equation. I do not understand why it is expecting 4 columns. Also to note the variables that are in my above snippet that are not explicitly defined are defined as [], serving as space holders.
Using fmincon is a huge overkill in this case. It's like using big heavy microscope to crack nuts... or martian rover to tow a pushcart or... anyway :) May be it's OK if you don't have to process large sets of vectors. If you need to fit hundreds of thousands of such vectors it can take hours. But this classic solution will be faster by several orders of magnitude.
%first make a n x 4 matrix of your vectors;
P=[W,X,Y,Z];
% now your equation looks like this S = P*m where m is 4 x 1 vectro
% containing your coefficients ( m = [a,b,c,d] )
% so solution will be simply
m_1 = inv(P'*P)*P'*S;
% or you can use this form
m_2 = (P'*P)\P'*S;
% or even simpler
m_3 = (S'/P')';
% all three solutions should give exactly same resul
% third solution is the neatest but may not work in every version of matlab
% Your modeled vector will be
Sm = P*m_3; %you can use any of m_1, m_2 or m_3;
% And your residual
R = S-Sm;
If you need to procees many vectors don't use for cycle. For cycles are very slow in Matlab and you should use matrices instead, if possible. S can also be nk matrix, where k is number vectors you want to process. In this case m will be 4k matrix.
What you are trying to do is similar to the answer I gave at is there a python (or matlab) function that achieves the minimum MSE between given set of output vector and calculated set of vector?.
The procedure is similar to what you are doing in EXCEL with solver. You create an objective function that takes (a, b, c, d) as the input parameters and output a measure of fit (mse) and then use fmincon or similar solver to get the best (a, b, c, d) that minimize this mse. See the code below (no MATLAB to test it but it should work).
function [a, b, c, d] = optimize_abcd(W, X, Y, Z)
%=========================================================
%Second argument is the starting point, second to the last argument is the lower bound
%to ensure the solutions are all positive
res = fmincon(#MSE, [0,0,0,0], [], [], [], [], [0,0,0,0], []);
a=res(1);
b=res(2);
c=res(3);
d=res(4);
function mse = MSE(x_)
a_=x_(1);
b_=x_(2);
c_=x_(3);
d_=x_(4);
S_ = a_*W + b_*X + c_*Y + d_*Z
mse = norm(S_-S);
end
end
I'm trying to solve the following equation for the variable h:
F is the normal cumulative distribution function
f is the probability density function of chi-square distribution.
I want to solve it numerically using Matlab.
First I tried to solve the problem with a simpler version of function F and f.
Then, I tried to solve the problem as below:
n = 3; % n0 in the equation
k = 2;
P = 0.99; % P* in the equation
F = #(x,y,h) normcdf(h./sqrt((n-1)*(1./x+1./y)));
g1 = #(y,h)integral(#(x) F(x,y,h).*chi2pdf(x,n),0,Inf, 'ArrayValued', true);
g2 = #(h) integral(#(y) (g1(y,h).^(k-1)).*chi2pdf(y,n),0,Inf)-P;
bsolve = fzero(g2,0)
I obtained an answer of 5.1496. The problem is that this answer seems wrong.
There is a paper which provides a table of h values obtained by solving the above equation. This paper was published in 1984, when the computer power was not good enough to solve the equation quickly. They solved it with various values:
n0 = 3:20, 30:10:50
k = 2:10
P* = 0.9, 0.95 and 0.99
They provide the value of h in each case.
Now I'm trying to solve this equation and get the value of h with different values of n0, k and P* (for example n0=25, k=12 and P*=0.975), where the paper does not provide the value of h.
The Matlab code above gives me different values of h than the paper.
For example:
n0=3, k=2 and P*=0.99: my code gives h = 5.1496, the paper gives h = 10.276.
n0=10, k=4 and P*=0.9: my code gives h = 2.7262, the paper gives h = 2.912.
The author says
The integrals were evaluated using 32 point Gauss-Laguerre numerical quadrature. This was accomplished by evaluating the inner integral at the appropriate 32 values of y which were available from the IBM subroutine QA32. The inner integral was also evaluated using 32 point Gauss-Laguerre numerical quadrature. Each of the 32 values of the inner integral was raised to the k-1 power and multiplied by the appropriate constant.
Matlab seems to use the same method of quadrature:
q = integral(fun,xmin,xmax) numerically integrates function fun from xmin to xmax using global adaptive quadrature and default error tolerances.
Does any one have any idea which results are correct? Either I have made some mistake(s) in the code, or the paper could be wrong - possibly because the author used only 32 values in estimating the integrals?
This does work, but the chi-squared distribution has (n-1) degrees of freedom, not n as in the code posted. If that's fixed then the Rinott's constant values agree with literature. Or at least, the literature that isn't behind a paywall. Can't speak to the 1984 Wilcox.
I have some data points to which I need to fit an exponential curve of the form
y = B * exp(A/x)
(without the help of Curve Fitting Toolbox).
What I have tried so far to linearize the model by applying log, which results in
log(y/B) = A/x
log(y) = A/x + log(B)
I can then write it in the form
Y = AX + B
Now, if I neglect B, then I am able to solve it with
A = pseudoinverse (X) * Y
but I am stuck with values of B...
Fitting a curve of the form
y = b * exp(a / x)
to some data points (xi, yi) in the least-squares sense is difficult. You cannot use linear least-squares for that, because the model parameters (a and b) do not appear in an affine manner in the equation. Unless you're ready to use some nonlinear-least-squares method, an alternative approach is to modify the optimization problem so that the modified problem can be solved using linear least squares (this process is sometimes called "data linearization"). Let's do that.
Under the assumption that b and the yi's be positive, you can apply the natural logarithm to both sides of the equations:
log(y) = log(b) + a / x
or
a / x + log(b) = log(y)
By introducing a new parameter b2, defined as log(b), it becomes evident that parameters a and b2 appear in a linear (affine, really) manner in the new equation:
a / x + b2 = log(y)
Therefore, you can compute the optimal values of those parameters using least squares; all you have left to do is construct the right linear system and then solve it using MATLAB's backslash operator:
A = [1 ./ x, ones(size(x))];
B = log(y);
params_ls = A \ B;
(I'm assuming x and y are column vectors, here.)
Then, the optimal values (in the least-squares sense) for the modified problem are given by:
a_ls = params_ls(1);
b_ls = exp(params_ls(2));
Although those values are not, in general, optimal for the original problem, they are often "good enough" in practice. If needed, you can always use them as initial guesses for some iterative nonlinear-least-squares method.
Doing the log transform then using linear regression should do it. Wikipedia has a nice section on how to do this:
http://en.wikipedia.org/wiki/Linear_least_squares_%28mathematics%29#The_general_problem
%MATLAB code for finding the best fit line using least squares method
x=input('enter a') %input in the form of matrix, rows contain points
a=[1,x(1,1);1,x(2,1);1,x(3,1)] %forming A of Ax=b
b=[x(1,2);x(2,2);x(3,2)] %forming b of Ax=b
yy=inv(transpose(a)*a)*transpose(a)*b %computing projection of matrix A on b, giving x
%plotting the best fit line
xx=linspace(1,10,50);
y=yy(1)+yy(2)*xx;
plot(xx,y)
%plotting the points(data) for which we found the best fit line
hold on
plot(x(2,1),x(2,2),'x')
hold on
plot(x(1,1),x(1,2),'x')
hold on
plot(x(3,1),x(3,2),'x')
hold off
I'm sure the code can be cleaned up, but that's the gist of it.
I have an unknown non-linear system and I want to model it using another system with some adaptable parameters (for instance, a neural network). So, I want to fix an online learning structure of the unknown system without knowing its dynamics, I can only interact with it through inputs-outputs. My problem is that I can not make it work in MATLAB using ode solvers. Lets say that we have this real system (my actual system is more complicated, but I will give a simple example in order to be understood):
function dx = realsystem(t, x)
u = 2;
dx = -3*x+6*u;
end
and we solve the equations like this:
[t,x_real] = ode15s(#(t,x)realsystem(t,x), [0 1], 0)
We suppose that is an unknown system and we do not know the coefficients 3 and 6 so we take an adaptive system with the 2 adaptive laws:
dx(t) = -p1(t)*x(t) + p2(t)*u(t)
dp1(t) = -e(t)*x(t)
dp2(t) = e(t)*u(t)
with e(t) the error e(t) = x(t) - x_real(t).
The thing is that I cannot find a way to feed the real values for each t to the ode solver in order to have online learning.
I tried with something like this but it didn't work:
function dx = adaptivesystem(t, x, x_real)
dx = zeros(3,1);
e = x_real - x;
u = 2;
dx(1) = -x(2)*x(1)+x(3)*u;
dx(2) = -e*x(1); %dx(2) = dp1(t)
dx(3) = e*u; %dx(3) = dp2(t)
end
You should be aware that your problem is ill-posed as it is. Given any trajectory x(t) obtained via sampling and smoothing/interpolating, you can choose p1(t) at will and set
p2(t) = ( x'(t) - p1(t)*x(t) ) / u.
So you have to formulate restrictions. One obvious is that the functions p1 and p2 should be valid for all trajectories of the black-box system. Do you have different trajectories available?
Another variant is to demand that p1 and p2 are constants. Actually, in this case and if you have equally spaced samples available, it would be easier to first find a good difference equation for the data. With the samples x[n] for time t[n]=t0+n*dt form a matrix X with rows
[ -u, x[n], x[n+1], ... ,x[n+k] ] for n=0, ... , N-k
and apply QR decomposition or SVD to X to determine the right hand kernel vectors. QR may fail to show a usable rank deficiency, so use the SVD on the top square part of R = USV^T, S diagonal, ordered as usual, U,V square and orthogonal, and use the last row of V, with coefficients
[b, a[0], ..., a[k] ],
corresponding to the smallest eigenvalue, to form the difference equation
a[0]*x[n]+a[1]*x[n-1]+...+a[k]*x[n-k]=b*u.
If the effective rank of R resp. S is not (k-1), then reduce k to be the effective rank plus one and start again.
If in the end k=1 is found, then you can make a differential equation out of it. Reformulate the difference equation as
a[0]*(x[n]-x[n-1])/dt = -(a[0]+a[1])/dt * x[n-1] + b/dt * u
and read off the differential equation
x'(t) = -(a[0]+a[1])/(a[0]*dt) * x(t) + b/(a[0]*dt) * u
One may reject this equation if the coefficients become uncomfortably large.
I'd like to find a way to fit a curve to a specific functional form, namely:
y=constant/x
Is there a good way to do this? My data is just a set of (x,y) pairs.
Sure, try this.
You can rewrite this equation as: y = c0 + c1*z where c0 and c1 are the constants you want to solve for and z = 1/x.
If you have n points, you can write one equation for each pair:
y1 = c0 + c1*z1
y2 = c0 + c1*z2
...
yn = c0 + c1*zn
You've got an (n x 1) vector of known y values on the left hand side. There's an (n x 2) matrix where the first column is all ones and the second column is the known vector of x values that multiplies a (2 x 1) vector of unknown coefficients c0 and c1.
Premultiply both sides by the (2 x n) transpose of the matrix and you'll have two equations for two unknown coefficients that you can solve easily.
Read this for details.
No reason to use anything more sophisticated than backslash, although the code inside backslash is quite sophisticated.
constant = (1./x(:))\y(:);
This does the linear regression for a model of the form y=constant/x. See that I inverted the elements of x using ./ and then applied backslash for the regression.
Do you have the optimization toolbox? If so, use the lsqcurvefit function.
a=lsqcurvefit(#(a,x) a(1)./x,1,x,y);
hold on
plot(x,y,'o') %plot data
plot(x,a./x) %fit